Tag: surface area and volume of cube and cuboid

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Questions Related to surface area and volume of cube and cuboid

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

$\angle B$ is a right angle is in $\Delta ABC$v and P,Q are points of trisection of hypotenuse $\bar{AC}.$ then $BP^{2}+BQ^{2}=\frac{5}{9}AC^{2}.$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Pythagorean theorem: In right triangle ABC, let BC be hypotenuse. If P, Q trisect AC, then AP = PQ = QC = AC/3. Using Pythagoras: BP^2 + BQ^2 = (AB^2 + AP^2) + (AB^2 + AQ^2) = 2AB^2 + (AC/3)^2 + (2AC/3)^2. With AB^2 + BC^2 = AC^2 for right triangle, this simplifies to 5/9 AC^2.

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

In $\triangle ABC$ right angled at $B, AB=5\ cm$ and $\angle ACB=30^{o}$ then the length of the sides $BC$ is

  1. $5\sqrt {3}$
  2. $2\sqrt {3}$
  3. $10\ cm$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given:- $ABC$ is a right angled triangle in which $AB = 5 \; cm$ and 


$\angle{ACB} = 30°$

Using trigonometric ratio,

$\tan{C} = \cfrac{AB}{BC}$

$\Rightarrow \tan{30°} = \cfrac{5}{BC}$

$\Rightarrow BC = 5 \sqrt{3} \; cm$

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

ABC is a triangle, right-angled at B. M is a point on BC. Hence,
$AM^{2}\, +\, BC^{2}\, =\, AD^{2}\, +\, BM^{2}$
State true or false.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $\angle ABC = 90$, M is a point of BC
In $\triangle ABM$, 
$AB^2 + BM^2 = AM^2$ (Pythagoras theorem)
$AB^2 = AM^2 - BM^2$ (I)

In $\triangle ABC$,
$AB^2 + BC^2 = AC^2$ (Pythagoras Theorem)
$AB^2 = AC^2 - BC^2$ (II)

Equating I and II,
$AM^2 - BM^2 = AC^2 - BC^2$
thus, $AM^2 + BC^2 = AC^2 + BM^2$

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

A guy wire attached to a vertical pole of height $18m$ is $24m$ long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

  1. $15.87m$
  2. $16.8m$
  3. $15$
  4. $15.67$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Consider the wire and pole system as a right angled triangle such that length of pole is the perpendicular, length of wire the hypotenuse and distance between the base of pole and the stack is the base, such that, H = 24 m, P = 18 m
Now, from Pythagoras theorem,
$H^2 = P^2 +B^2$
$24^2 = 18^2 + B^2$
$576 = 324 + B^2$
$B^2 = 252$
$B = 15.87$ m
Thus, distance between the base of the pole and the stack is 15.87 m

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

A person wishes to fit three rods together in the shape of a right-angled triangle so that the hypotenuse is to be $4:cm$ longer than the base and $8:cm$ longer than the altitude. The lengths of the rods are:

  1. $3\:cm$, $4\:cm$, $5\:cm$
  2. $1.5\:cm$, $2\:cm$, $2.5\:cm$
  3. $6\:cm$, $8\:cm$, $10\:cm$
  4. $12\:cm$, $16\:cm$, $20\:cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the altitude$=x:cm$
$\therefore$ The Base$=(x+4):cm$
and the Hypotenuse$=(x+8):cm$

Using Pythtegores Theorem
$(x+8)^2=(x+4)^2+x^2$
$x^2-8x-48=0$
$(x-12)(x+4)=0$
$x=12$
$\therefore$ The sides are $12:cm$, $16:cm$, $20:cm$ 

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

What is the value of the hypotenuse of a right triangle whose sides are $12$ and $18$?

  1. $4.24$
  2. $3.46$
  3. $2.16$
  4. $21.63$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the sides be $a=12$ and $b=18$

Le the hypotenuse be $c$
Using Pythagoras theorem
${ c }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }\ { c }^{ 2 }={ (12) }^{ 2 }+{ (18) }^{ 2 }\ { c }^{ 2 }=144+324\ { c }^{ 2 }=468\ c=\sqrt { 468 } \ c=21.63$

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

A triangle whose lengths of sides are $5$ cm, $12$ cm and $13$ cm. The triangle is ____________.

  1. Obtuse-angled triangle

  2. Acute-angled triangle

  3. Right-angled triangle

  4. Triangle is not formed
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given sides of triangle are $5$ cm, $12$ cm, $13$ cm
Now applying pythagorus theorem:
$h^{2}= b^{2}+P^{2}$
$\Rightarrow 13^{2}= 12^{2}+5^{2}$
$ \Rightarrow 169 =144+25$
$ \Rightarrow 169 =169$
These three sides clearly satisfy ptyhagorous theorem.
Therefore, the triangle is RIGHT ANGLED triangle.

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

In any triangle $ABC$,  $AB^{2} + AC^{2} = 3 (AO^{2} + OC^{2})$.
where $O$ is mid-point of $BC$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By Apollonius theorem, AB^2 + AC^2 = 2(AO^2 + BO^2). Since O is the midpoint of BC, BO = OC. Thus, AB^2 + AC^2 = 2(AO^2 + OC^2). The given equation 3(AO^2 + OC^2) is incorrect.

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

Two cars are travelling along two roads which cross each other at right angles at $A$. One car is travelling towards A at $21\ kmph $ and the other is travelling towards $A$ at $28\ kmph$.If initially, their distance from $A$ are $1500\ km$ and $2100\ km$ respectively,then the nearest distance between them is ,

  1. $30$
  2. $45$
  3. $60$
  4. $75$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Nearest distance
$=\dfrac{2100-1500}{10}$
$\dfrac{600}{10}=60$